If vec{a}=hat{i}+hat{j}+hat{k},vec{b}=-hat{i} | vedclass

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(D) Using the vector triple product identity $(\vec{u} 	imes \vec{v}) 	imes \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u

Vedclass · Accepted Answer

$-12$

Vedclass · Answer

(D) Using the vector triple product identity $(\vec{u} 	imes \vec{v}) 	imes \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$,let $\vec{u} = \vec{a}$,$\vec{v} = \vec{b}$,and $\vec{w} = (\vec{a} 	imes \vec{c})$.Then $(\vec{a} 	imes \vec{b}) 	imes (\vec{a} 	imes \vec{c}) = [\vec{a} \vec{b} (\vec{a} 	imes \vec{c})] \vec{a} - [\vec{b} (\vec{a} 	imes \vec{c}) \vec{a}] \vec{a} = [\vec{a} \vec{b} \vec{c}] \vec{a} - 0 = [\vec{a} \vec{b} \vec{c}] \vec{a}$.Now,$(\vec{a}-\vec{b}) \cdot [(\vec{a} 	imes \vec{b}) 	imes (\vec{a} 	imes \vec{c})] = (\vec{a}-\vec{b}) \cdot ((\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}) = [\vec{a} \vec{b} \vec{c}] (\vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{a})$.Calculating the scalar triple product $[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & 1 & 1 \ -1 & 2 & -2 \ 2 & -1 & 2 \end{vmatrix} = 1(4-2) - 1(-2+4) + 1(1-4) = 2 - 2 - 3 = -3$.Calculating $(\vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{a}) = (3) - (-1+2-2) = 3 - (-1) = 4$.Thus,the result is $(-3) 	imes 4 = -12$.

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=-\hat{i}+2\hat{j}-2\hat{k}$ and $\vec{c}=2\hat{i}-\hat{j}+2\hat{k}$,then $(\vec{a}-\vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c})]$ is

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